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Full text of "Treatise On Analysis Vol-Ii"

9   THE PLANCHEREL-GODEMENT THEOREM       381

The Plancherel-Godement theorem applies in particular when the
bitrace g is of the form (x, y) \-+f(y*x), where /is a positive linear form on A
(and therefore a trace since A is commutative). When g comes from a trace/
the formula (15.9.2.1) leads us to ask whether we also have

(15.9.3)

= f

JS

A partial answer to this question is provided by the following theorem:

(15.9.4) (Bochner-Godement theorem) (i) Let f be a positive linear
form on a commutative algebra A with involution, such that the bitrace
0(x, y) = f(xy*) satisfies the hypotheses of (15.9.2). Then, if the formula
(15.9.3) is true and if the measure mg is bounded, f satisfies the condition

(B)    There exists a real number M > 0 such that \f(x)\2 £ M -f(xx*)for
all x e A.

(ii) Conversely, let f be a positive linear form on A which satisfies the
condition (B), and suppose that the corresponding bitrace g(x, y) — f(xy*)
satisfies (U) and is such that the prehilbert space A/ng and the star algebra $0g
are separable. Then g also satisfies (N), the measure mg is bounded, and the
formula (15.9.3) is valid,

(i) If mg is bounded and if (15.9.3) holds, then the Cauchy-Schwarz
inequality (13.11.2.2) gives us

f*dm/^ro,(S,)- f

and hence the inequality of (B), with M =

(ii) Recall that the definition of the Hilbert space Ug does not pre-
suppose that the condition (N) is satisfied. The inequality in (B) can be
put in the form \f(x)\2 ^ M\\ng(x)\\2, and shows that / vanishes on n^;
hence we may write f=f'°ng, where /' is a linear form on A/ng. Also
\f'(n9(x))\2 = M ||7T0(x)||2, so that/' is continuous on the prehilbert space
A/110 (5.5.1) and therefore extends to a continuous linear form on the Hilbert
space H^ (5.5.4). Hence, by (6.4.2), there exists a well-determined vector
a e H^ such that

(15.9.4.1)                               f(x) = (n,(x)\d).

From this it follows that, for all # e A, we have

(15.9.4.2)                                V9(x)-a = n,(x). \x(y)$(l) dm(%). The argument of